The initial-boundary value problem for the 1D nonlinear Schrödinger equation on the half-line

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1 The initial-boundary value problem for the 1D nonlinear Schrödinger equation on the half-line Justin Holmer Department of Mathematics, University of California Berkeley, CA Abstract We prove, by adapting the method of Colliander-Kenig [9], local wellposedness of the initial-boundary value problem for the one-dimensional nonlinear Schrödinger equation i t u + xu 2 + λu u α 1 = 0 on the half-line under low boundary regularity assumptions. 1 AMS subject code: 35Q55 2 The content of this article appears as part of the author s Ph.D. thesis at the University of Chicago. 1

2 1 Introduction We consider the initial-boundary value problem on the right half-line for the one-dimensional nonlinear Schrödinger (1D NLS) equation i t u + xu 2 + λu u α 1 = 0 for (x, t) (0, + ) (0, T ) u(0, t) = f(t) for t (0, T ) (1) u(x, 0) = φ(x) for x (0, + ) where λ C. On R, we define the homogeneous L 2 -based Sobolev spaces Ḣs = Ḣ(R) by the norm φ Ḣs = ξ s ˆφ(ξ) L 2 ξ and the L 2 -based inhomogeneous Sobolev spaces H s = H s (R) by the norm φ H s = ξ s ˆφ(ξ) L 2 ξ, where ξ = (1 + ξ 2 ) 1/2. In addition, we shall need L 2 -based inhomogeneous Sobolev spaces on the half-line R + = (0, + ), which we denote H s (R + ). These are defined, for s 0, as: φ H s (R + ) if φ H s (R) such that φ(x) = φ(x) for a.e. x > 0; in this case we set φ H s (R + ) = inf φ φ H s (R). We also similarly define, for s 0, φ H s (0, L) if φ H s (R) such that φ(x) = φ(x) a.e. on (0, L); in this case we set φ H s (0,L) = inf φ φ H s. The local smoothing inequality of [15] for the 1D Schrödinger group is e it 2 x φ L x Ḣ 2s+1 t c φ Ḣs This inequality is sharp in the sense that 2s+1 cannot be replaced by any higher number. We are thus motivated to consider initial-boundary data pairs (φ(x), f(t)) H s (R + x ) H 2s+1 (R + t ) and inclined to consider this configuration optimal in the scale of L 2 -based Sobolev spaces. Note that the trace map φ φ(0) is well-defined on H s (R + ) when s >

3 Thus, if s > 1 2s+1, then > 1 and both φ(0) and f(0) are well-defined quantities. 2 2 Since φ(0) and f(0) are both meant to represent u(0, 0), they must agree. Therefore, we consider (1) for 0 s < 3 2 in the setting φ H s (R + ), f H 2s+1 (R + ), and if 1 < s < 3, φ(0) = f(0) (2) 2 2 The solutions we construct shall have the following characteristics. Definition 1. u(x, t) will be called a distributional solution of (1), (2) on [0, T ) with strong traces if (a) u belongs to a space X with the property that u X implies u u α 1 is defined as a distribution. (b) u(x, t) satisfies the equation (1) in the sense of distributions on the set (x, t) (0, + ) (0, T ). (c) Space traces: T < T, we have u C([0, T ]; H s x) and u(, 0) = φ in H s (R + ). (d) Time traces: T < T, we have u C(R x ; H 2s+1 (0, T )) and u(0, ) = f in H 2s+1 (0, T ). For the purposes of uniqueness in the high regularity setting s > 1, we can 2 consider a weaker notion of solution. Definition 2. u(x, t) will be called a distributional solution of (1), (2) on [0, T ) with weak traces if it satisfies conditions (a), (b) of Definition 1 and (c) One-sided space traces: T < T, we have u C([0, T ]; H s (R + x )) and u(, 0) = φ in H s (R + ). 3

4 (d) Boundary values: T < T, we have lim x 0 u(x, ) f H 2s+1 (0,T ) = 0. So that we may, at a later time, properly address the matter of uniqueness in the low regularity s < 1 2 used by [1]. setting, we shall introduce the concept of mild solution Definition 3. u(x, t) is a mild solution of (1) on [0, T ) if T < T, a sequence {u n } in C([0, T ]; H 2 (R + x )) C 1 ([0, T ]; L 2 (R + x )) such that (a) u n (x, t) solves (1) in L 2 (R + x ) for 0 < t < T. (b) (c) lim u n u n + C([0,T ]; H s (R + x )) = 0. lim n + u n(0, ) f H 2s+1 (0,T ) = 0. [1] have announceed a method for proving uniqueness of mild solutions for the Korteweg-de Vries (KdV) equation on the half-line (to be discussed further in [2]), and the techniques of this forthcoming paper may also apply here to resolve the uniqueness problem for 0 s < 1 2. We establish in 8 the following straightforward fact. Proposition 1. For s > 1, u is a distributional solution of (1), (2) with weak 2 traces if and only if it is a mild solution; in this case u is unique. Our main result is the following existence statement. Theorem 1. (a) Subcritical: Suppose 0 s < 1 5 2s, and 2 α < 2 1 2s

5 or 1 < s < 3, and 2 α < 2 2 Then T > 0 and u that is both a mild solution and a distributional solution with strong traces of (1),(2) on [0, T ). If T <, then lim t T u(, t) H s x =. Also, T < T, δ 0 = δ 0 (s, T, φ, f) > 0 such that if 0 < δ δ 0 and φ φ 1 H s (R + ) + f f 1 2s+1 < δ then there is a solution u H 1 (as (R + ) above) on [0, T ], corresponding to (φ 1, f 1 ), such that u u 1 C([0,T ]; H s x ) + u u 1 C(Rx; H 2s+1 cδ, with c = c(s, T, f, φ). (0,T )) (b) Critical: Suppose 0 s < 1 2 and α = 5 2s 1 2s. Then T > 0 maximal and u that is both a mild solution and a distributional solution with strong traces of (1),(2) on [0, T ). Also, T = T (s, φ, f) < T and δ 0 = δ 0 (s, φ, f) > 0 such that if 0 < δ δ 0 and φ φ 1 H s (R + ) + f f 1 H 2s+1 (R + ) < δ then there is a solution u 1 (as above) on [0, T ], corresponding to (φ 1, f 1 ), such that u u 1 C([0,T ]; H s x ) + u u 1 C(Rx; H 2s+1 cδ, with c = c(s, f, φ). (0,T )) Note that in (b), we may not have blow-up in the norm u(, t) as t T. The proof of Theorem 1 involves the introduction of a boundary forcing operator analogous to that introduced by [9] in their treatment of the generalized Korteweg de-vries equation (gkdv) on the half-line, and incorporates the techniques of the standard proof of local well-posedness of the corresponding initial-value problem based on the Strichartz estimates (see [7]). One could also consider the left half-line problem i t u + xu 2 + λu u α 1 = 0 for (x, t) (, 0) (0, T ) u(0, t) = f(t) for t (0, T ) u(x, 0) = φ(x) for x (, 0) 5

6 although this is actually identical to the right half-line problem (1) by the transformation u(x, t) u( x, t). We plan, in a future publication, to examine the initial-boundary value problem for the line-segment i t u + xu 2 + λu u α 1 = 0 for (x, t) (0, L) (0, T ) u(0, t) = f 1 (t) for t (0, T ) u(l, t) = f 2 (t) for t (0, T ) u(x, 0) = φ(x) for x (0, L) and consider global existence questions for the half-line and line-segment problems. We now briefly mention some earlier work and alternate perspectives on this problem and related problems. The main new feature of our work is the low regularity requirements for φ and f. Under higher regularity assumptions, more general results are already available. [18] considered a bounded or unbounded domain Ω R n with smooth boundary Ω, and proved global existence of solutions to i t u + u + λu u α 1 = 0 for (x, t) Ω (0, T ) u(x, t) = f(x, t) for x Ω (3) u(x, 0) = φ(x) for x Ω where f C 3 ( Ω) is compactly supported, φ H 1 (Ω), and λ < 0. This solution is obtained as a limit of solutions to approximate problems after several a priori identities have been established. Earlier, [6] and [5] had obtained solutions to (1) for α > 3, λ < 0 and α = 3, λ R for φ H 2 (R + ) and f C 2 (0, T ), using 6

7 semigroup techniques and a priori estimates. The problem (3) with f = 0 had been considered previously ([] [22] [20] [21] [23]). [10] in the integrable case α = 3, λ = ±2 with φ Schwartz and f sufficiently smooth, obtained a solution to (1) by reformulating the problem as a 2 2 matrix Riemann-Hilbert problem. In this setting, [3] obtain an explicit representation for x u(0, t). Outline: In 2, we discuss some notation, introduce function spaces and recall some needed properties of these function spaces. In 3, we review the definition and basic properties of the Riemann-Liouville fractional integral. In, 5, we state the needed estimates for the group and inhomogeneous solution operator. In 6, we define the boundary forcing operator, adapted from [9], and prove the needed estimates for it. In 7, we prove Theorem 1. In 8, we prove Prop Notations and some function space properties Let χ S denote the characteristic function for the set S. We shall write L q T to mean L q ([0, T ]). Set ˆφ(ξ) = x e ixξ φ(x) dx. Define (τ i0) α as the limit, in the sense of distributions, of (τ + iγ) α as γ 0. Let ξ s = (1 + ξ 2 ) s/2. Let D s f(ξ) = ξ s ˆf(ξ). The homogeneous L 2 -based Sobolev spaces are Ḣs (R) = ( 2 ) s/2 L 2 (R) and the inhomogeneous L 2 -based H s (R) = (1 2 ) s/2 L 2 (R). We also set, for 1 p, W s,p = (I 2 ) s/2 L p. We use the notation H s to mean H s (R) (and not H s (R + ) or H0(R s + )). The trace operator φ φ(0) is defined for φ H s (R) when s > 1. For s 0, define φ 2 Hs (R + ) if φ H s (R) 7

8 such that φ(x) = φ(x) for x > 0; in this case we set φ H s (R + ) = inf φ φ H s (R). For s 0, define φ H0(R s + ) if, when φ(x) is extended to φ(x) on R by setting φ(x) = 0 for x < 0, then φ H s (R); in this case we set φ H s 0 (R + ) = φ H s (R). Define φ C0 (R + ) if φ C (R) with supp φ [0, + ) (so that, in particular, φ and all of its derivatives vanish at 0), and C0,c(R + ) as those members of C0 (R + ) with compact support. We remark that C0,c(R + ) is dense in H0(R s + ) for all s R. We shall take a fixed θ Cc (R) such that θ(t) = 1 on [ 1, 1] and supp θ [ 2, 2]. Denote by θ T (t) = θ(tt 1 ). Lemma 1 ([9] Lemma 2.8). If 0 α < 1 2, then θ T h H α where c = c(α, θ). c T α h Ḣα, Lemma 2 ([13] Lemma 3.5). If 1 2 < α < 1 2, then χ (0,+ )f H α c f H α, where c = c(α). Lemma 3 ([9] Prop. 2., [13] Lemma 3.7, 3.8). If 1 2 < α < 3 2, then H α 0 (R + ) = {f H α (R + ) f(0) = 0} and if f H α (R + ) with f(0) = 0, then χ (0,+ ) f H α 0 (R + ) c f H α (R + ), where c = c(α). The following Gronwall-type inequality can be obtained by applying the Hölder inequality iteratively: Lemma. If 1 q 1 < q and t 0 ( t 0 1/q ( t ) 1/q1 g(s) ds) q cδ + c f(s) q 1 ds 0 then with γ defined by 2cγ 1 q 1 1 q = 1, we have t 0, ( t 1/q1 f(s) q 1 ds) (γt) γt δ 0 8

9 A version of the chain rule for fractional derivatives is Lemma 5 (Prop. 3.1 in [8]). Let 0 < s < 1, u : R R 2 and F : R 2 R 2, F C 1, so that F (u) is a 2 2 matrix. Then D s F (u) L r c F (u) L r 1 D s u L r 2 for 1 r = 1 r r 2 with 1 < r, r 1, r 2 <. The product rule for fractional derivatives is Lemma 6 (Prop. 3.3 in [8]). Let 0 < s < 1. If u, v : R R, then D s (uv) L r D s u L r 1 v L r 2 + u L r 3 D s v L r for 1 < r, r 1, r 2, r 3, r < and 1 r = 1 r r 2, 1 r = 1 r r. 3 The Riemann-Liouville fractional integral The tempered distribution tα 1 + Γ(α) Re α > 0, i.e. t α 1 + Γ(α), f = 1 Γ(α) is defined as a locally integrable function for + Integration by parts gives, for Re α > 0, that t α 1 + Γ(α) = k t 0 [ t α+k 1 ] + Γ(α + k) t α 1 f(t) dt for all k N. This formula can be used to extend the definition (in the sense of distributions) of tα 1 + Γ(α) to all α C. In particular, we obtain t α 1 + Γ(α) = δ 0 (t) α=0 9

10 A change of contour calculation shows that [ t α 1 + ] (τ) = e 1 2 πiα (τ i0) α Γ(α) where (τ i0) α is the distributional limit. If f C 0 (R + ), we define Thus, when Re α > 0, I α f(t) = 1 Γ(α) I α f = tα 1 + Γ(α) f t 0 (t s) α 1 f(s) ds and I 0 f = f, I 1 f(t) = t 0 f(s) ds, and I 1f = f. Also I α I β = I α+β, which follows from the Fourier transform formula. For further details on the distribution t α 1 +, see [11]. Γ(α) Lemma 7. If h C 0 (R + ), then I α h C 0 (R + ), for all α C. Lemma 8 ([12]). If 0 α < + and s R, then I α h H s 0 (R + ) c h H s+α 0 (R + ) Lemma 9 ([12]). If 0 α < +, s R, µ C 0 (R) where c = c(µ). µi α h H s 0 (R + ) c h H s α 0 (R + ) Estimates for the group Set 1 e it 2 x φ(x) = e ixξ e itξ2 ˆφ(ξ) dξ () 2π ξ 10

11 so that (i t + x)e 2 it 2 x φ = 0 e it 2 x φ(x) t=0 = φ(x) for (x, t) R R for x R Lemma 10. Let s R. If φ H s (R), then (a) Space traces: e it 2 x φ(x) C(Rt; H s x) c φ H s. (b) Time traces: θ T (t)e it 2 x φ(x) 2s+1 c T 1/ φ C(R x; H H s. t ) (c) Mixed-norm: If 2 q, r and = 1, then q 2r eit 2 x φ(x) L q t W x s,r c φ H s. Proof. (a) is clear from (). (b) was obtained in [15]. (c) was obtained by [19] (see also [1]). 5 Estimates for the Duhamel inhomogeneous solution operator Let Then Dw(x, t) = i t 0 e i(t t ) 2 x w(x, t ) dt (i t + x)dw(x, 2 t) = w(x, t) for (x, t) R R Dw(x, 0) = 0 for x R Lemma 11. Suppose 2 q, r and 1 q + 1 2r = 1, then (a) Space Traces: If s R, then Dw C(Rt;H s x) c w L q (b) Time Traces: If 3 2 < s < 1 2, then θ T (t)dw(x, t) 11 t W x s,r. 2s+1 C(R x; H t ) c T 1/ w L q t W x s,r.

12 (c) Mixed-norm: If s R, then Dw L q t W s,r x c w L q t W x s,r Proof. (a) and (c) are due to [19] (see also [1]). We now prove (b), following the techniques of Theorem 2.3 in [16]. We use the representation Dw(x, t) = i πi = I + II + and Term II can also be written τ (sgn t )e i(t t ) x 2 w(x, t ) dt [ ] e itτ 1 ixξ ŵ(ξ, τ) lim e ɛ 0 + 2π τ + ξ dξ dτ 2 τ+ξ 2 >ɛ II = 1 e itτ [m(, τ) ŵ t (, τ)](x) dτ 2π τ where ŵ t (, τ) denotes the Fourier transform of w(x, t) in the t-variable alone and m(x, τ) = 1 2 χ (0,+ )(τ) exp( x τ 1/2 ) + 1 τ 1/2 2 χ (,0)(τ) sin( x τ 1/2 ) τ 1/2 First we treat Term I for all s and all admissible pairs q, r. Pairing Term I with f(x, t) such that f L 1 xh 2s+1 t and 1, we are left to show that H (sgn t it 2 )e x w(x, t ) dt t s x H θ T (t)e it 2x f(x, t) dt t s x c w L q t W x s,r c f L 1 xh 2s+1 t The first of these follows from the proof of (a), while the second is obtained by duality and Lemma 10(b). We address Term II separately for r = 2, q = 1, and r = 1, q = ; the intermediate cases follow by interpolation. For the case 3 12.

13 r = 2, q = 1, we use the first representation of Term II with Lemma 1, the change of variable η = ξ 2, and L 2 -boundedness of the Hilbert transform on A 2 -weighted spaces, to obtain θ T (t)(term II) 2s+1 H t ( c c ( ξ ξ ) 1/2 ξ s ŵ(ξ, τ) 2 dξ ξ s ( t ŵ x (ξ, t) dt) 2 dξ ) 1/2 where ŵ x (ξ, t) denotes the Fourier transform in the x-variable alone. Complete the bound by applying Minkowskii s integral inequality and the Placherel theorem. The validity of this step is restricted to 3 2 < s < 1 2. We shall only prove the r = 1, q = 3 case for s = 0. Note that by the second representation for Term II, (Term II) L is x H 1/ t τ 1/2 m(x y, τ)ŵ t (y, τ) dy τ 1/2 m(x z, τ)ŵ t (z, τ) dz dτ τ y which is equivalent to where y,s,z,t K(y, s, z, t) = z K(y, s, z, t)w(y, s)w(z, t) dy ds dz dt τ τ 1/2 e i(s t)τ m(x y, τ)m(x z, τ) dτ From the definition of m, we see that K(y, s, z, t) c s t 1/2. We conclude by applying the theorem on fractional integration (see Theorem 1 of Chapter V in [17]). 13

14 6 Estimates for the Duhamel boundary forcing operator For f C 0 (R + ), define the boundary forcing operator t Lf(x, t) = 2e i 1 π e i(t t ) x 2 δ0 (x)i 1/2 f(t ) dt (5) 0 = 1 t ( ) ix (t t ) 1/2 2 exp I π (t t 1/2 f(t ) dt (6) ) 0 The equivalence of the two definitions is evident from the formula [ ( e i π sgn t 2 1 ix 2 exp )] (ξ) π t 1/2 = e itξ 2 t From these two definitions, we see that (i t + x)lf(x, 2 t) = 2e i 3 π δ 0 (x)i 1/2 f(t) Lf(x, 0) = 0 Lf(0, t) = f(t) for (x, t) R R for x R for t R We now establish some continuity properties of Lf(x, t) when f is suitably nice. Lemma 12. Let f C0,c(R + ). (a) For fixed t, Lf(x, t) is continuous in x for all x R and x Lf(x, t) is continuous in x for x 0 with lim x Lf(x, t) = e 1 πi I 1/2 f(t) x 0 lim x Lf(x, t) = e 1 πi I 1/2 f(t) x 0 (7) (b) k = 0, 1, 2,... and for fixed x, k t Lf(x, t) is continuous in t for all t R. 1

15 We also have the pointwise estimates, for k = 0, 1, 2,..., on [0, T ], k t Lf(x, t) + x Lf(x, t) c x N where c = c(f, N, k, T ). Proof. Let us denote integration by parts by IBP. It is clear from (6) and dominated convergence that, for fixed t, Lf(x, t) is continuous in x, and for fixed x, Lf(x, t) is continuous in t. Let h = 2e i 1 π I 1/2 f C 0 (R + ) (by Lemma 7) and φ(ξ, t) = t 0 e i(t t )ξ h(t ) dt. By IBP in t, k ξ φ(ξ, t) c ξ k 1, where c = c(h, k, T ), and thus k ξ φ(ξ 2, t) c ξ k 2 (8) We have Lf(x, t) = e ixξ φ(ξ 2, t) dξ (9) ξ and by IBP in ξ and (8), we have Lf(x, t) c x N. By t [e i(t t ) 2 x δ0 (x)] = t [e i(t t ) 2 x δ0 (x)] and IBP in t in (5), t Lf = L t f, and thus, for fixed x, k t Lf(x, t) is continuous in t and k t Lf(x, t) c x N. By 2 x[e i(t t ) 2 x δ0 (x)] = i t [e i(t t ) 2 x δ0 (x)] and IBP in t in (5), 2 xlf(x, t) = 2e i 3 π δ 0 (x)i 1/2 f(t) il( t f)(x, t). Hence x x Lf(x, t) = e i 3 π (sgn x)i 1/2 f(t) i L( t f)(x, t) dx + c(t) x =0 Since all terms except c(t) are odd in x, we must have c(t) = 0. From this we obtain (7), and the bound x Lf(x, t) c. From (9), IBP in ξ and (8), we obtain that x Lf(x, t) c x N. Combining the two previous bounds, we have x Lf(x, t) c x N. Now we provide an alternate representation of Lf(x, t). 15

16 Lemma 13. Suppose f C 0,c(R + ). Then where Lf(x, t) = 1 e itτ e x (τ i0)1/2 ˆf(τ) dτ (10) 2π τ (τ i0) 1 2 = χ(0,+ ) (τ) τ 1/2 iχ (,0) (τ) τ 1/2 Proof. It suffices to verify that (a) On [0, T ], Lf(x, t) + t Lf(x, t) c x N, with c = c(f, N, T ). (b) Lf(x, 0) = 0 (c) (i t + 2 x)lf(x, t) = 2δ 0 (x)e 3 πi I 1/2 f(t) (a) is integration by parts in τ in (10) using 2(τ i0) 1/2 x 1 τ [e x (τ i0)1/2 ] = e x (τ i0)1/2. To show (b), note that since f C 0,c(R + ), ˆf(τ) extends to an analytic function on Im τ < 0 satisfying ˆf(τ) c τ k with c = c(f, k), and thus Lf(x, 0) = 1 2π lim e x τ 1/2 ˆf(τ) dτ (11) γ 0 Im τ=γ Since e x τ 1/2 1 for Im τ < 0, by Cauchy s theorem, (11) = 0. (c) is a direct computation from (10). Denote the operator defined by (10) as L 2 f(x, t) and the one given by (5)-(6) as L 1 f(x, t). Setting w = L 1 f L 2 f, we have w(x, 0) = 0 and (i t + x)w 2 = 0. Compute t x w 2 dx = 0, which yields w = 0, to complete the proof. Lemma 1. Suppose q, r 2 and 1 q + 1 2r = 1. (a) Space traces: If 1 2 < s < 3 2, then θ T (t)lf(x, t) C(Rt;H s x) c T 1/ f 2s+1 H 0 (R + ). 16

17 (b) Time traces: If s R, then Lf C(R x;h 2s+1 0 (R + t )) c f H (c) Mixed norm: If 0 s 1, r, we have Lf L q t W s,r 2s+1 0 (R + ) x c f H Proof. By density, it suffices to establish these facts for f C 0,c(R + ). show But By pairing (a) with φ(x) such that φ H s t t =0. 2s+1 0 (R + ) 1, we see that it suffices to f(t )θ T (t)e i(t t ) 2 x φ x=0 dt c T 1/ f H 2s+1 LHS χ (,t) f(t ) 2s+1 θ H T (t)e i(t t ) 2 x φ(x) 2s 1 RHS t H t by Lemmas 10(b) and 2. To establish the continuity statement, write θ T (t 2 )Lf(x, t 2 ) θ T (t 1 )Lf(x, t 1 ) = t 2 t 1 t [θ(t)lf(x, t)] dt. By t L = L t and the bound just derived, we have θ T (t 2 )Lf(x, t 2 ) θ T (t 1 )Lf(x, t 1 ) c t 2 t 1 f 2s+5 H 0 (b) is immediate from Lemma 13, except that we should confirm that (under the assumption f C 0,c(R + ), that k t Lf(x, 0) = 0 for all k = 0, 1, 2,.... This, however, follows from t L = L t. The continuity statement follows by using Lf(x 2, t) Lf(x 1, t) = x 2 x 1 and thus x Lf(x, t) = e 1 πi (sgn x) 1 2π Lf(x 2, t) Lf(x 1, t) To prove (c), it suffices to establish x Lf(x, t) dx. From Lemma 13, we have e itτ e x (τ i0)1/2 [I 1/2 f] (τ) dτ τ 2s+1 H 0 (R + ) c x 2 x 1 f 2s+3 H 0 Lf(x, t) L t L c f x Ḣ1/ (12).. and x Lf(x, t) L t L c f x Ḣ3/ (13) 17

18 Indeed, the proof of (a) in the case s = 1 shows Lf(x, t) L t L 2 x c f Ḣ 1/ xlf(x, t) L t L 2 x c f Ḣ 3/ Interpolate (12) with the first inequality and (13) with the second inequality to obtain Lf(x, t) L q t Lr x c f Ḣ 1/ xlf(x, t) L q t Lr x c f Ḣ 3/ for admissible q, r. This implies Lf(x, t) L q t W x s,r c f 2s+1 H, r for s = 0 and s = 1. Now interpolate over s between these two endpoints to obtain the result as stated. By pairing LHS of (12) against w(x, t) L /3 t L 1 x, we see that it suffices to show e itτ e x (τ i0)1/2 w(x, t) dx dt Ḣ 1/ x t Writing out the L 2 τ norm, we see that it suffices to show K(x, t, y, s) w(x, t)w(y, s) dx dt dy ds c w /3 L t L 1 x where x,t,y,s K(x, t, y, s) = τ τ 1/2 e i(t s)τ e x (τ i0)1/2 e y (τ+i0)1/2 dτ By a change of contour calculation, it follows that K(x, y, t, s) c t s 1/2, and hence (12) follows by the theorem on fractional integration. For (13), the kernel is instead K(x, t, y, s) = (sgn x)(sgn y) and hence the estimation of K is identical. τ τ 1/2 e i(t s)τ e x (τ i0)1/2 e y (τ+i0)1/2 dτ 18

19 7 Existence: Proof of Theorem 1 First we prove the subcritical assertion (a) in the case 0 s < 1. Select an 2 extension φ H s of φ such that φ H s 2 φ H s (R + ). Set r = α+1 1+(α 1)s q = (α+1) 2. This is an admissible pair with r 2 and q 2( + 1). Set (α 1)(1 2s) 1 2s Z = C(R t ; Hx) s C(R x ; H 2s+1 t ) L q t Wx s,r Take w Z. By the chain rule (Lemma 5), for α 1 (see below for details) for some σ > 0. D s ( w α 1 w) L q T Lr x and ct σ w α L q T W r,s x (1) Note that by Lemmas 10(b), 11(b), 2, if w Z, then f(t) θ 2T (t)e it 2 x φ x=0 H 2s+1 0 (R + t ) and θ 2T (t)d(w w α 1 )(0, t) H 2s+1 0 (R + t ), and the evaluation at x = 0 in these statements is understood in the sense of C(R x ; H 2s+1 t ). Let Λw(t) = θ T (t)e it 2 x φ + θt (t)l(f θ 2T e i 2 x φ x=0 )(t) (15) λθ T (t)d(w w α 1 )(t) + λθ T (t)l(θ 2T D(w w α 1 ) x=0 )(t) so that, on [0, T ], (i t + 2 x)λw = λw w α 1 for x 0 in the sense of distributions. By Lemmas 10, 11, 1 and (1), Λw Z c φ H s (R + ) + c f H 2s+1 (R + ) + ct σ w α Z (16) In the sense of C(R t ; H s x), we have Λw ( x, 0) = φ(x) on R, and in the sense of C(R x ; H 2s+1 t ), we have Λw(0, t) = f(t) on [0, T ]. We therefore look to solve Λw = w for some selection of T. By the chain rule and product rule (see below for details), for α 2, Λw 1 Λw 2 Z ct σ ( w 1 α 1 Z T 19 + w 2 α 1 Z ) w 1 w 2 Z (17)

20 Now choose T small in terms of φ H s (R + ) and f 2s+1, so that, by (16) H (R + ) and (17), Λ is a contraction, which yields a unique fixed point u, which on [0, T ] solves the integral equation u(t) = e it 2 x φ + L(f e i 2 x φ x=0 ) (18) λd(u u α 1 ) + λl(d(u u α 1 ) x=0 ) Let S be the set of all times T > 0 for which (1) u Z such that u solves (18) on [0, T ] and (2) for each pair u 1, u 2 Z, such that u 1 solves (18) on [0, T 1 ] with T 1 T and u 2 solves (18) on [0, T 2 ] with T 2 T, we have u 1 = u 2 on [0, min(t 1, T 2 )]. We claim that T as given in the above contraction argument is in S. We need only show condition (2). But the integral equation (18) has a unique solution by the contraction argument in the space L q T m W s,r x, where T m = min(t 1, T 2 ), by Lemmas 10(c), 11(c), 1(c) and the fact that χ [0,Tm]Lg = χ [0,Tm]L(θ Tm g), χ [0,Tm]Dw = χ [0,Tm]Dχ [0,Tm]w. Let T = sup S. Define u on [0, T ) by setting, for t < T, u (t) = u(t) for some u Z whose existence is given by condition (1); this is well-defined by condition (2). Suppose T < and lim t T u(, t) H s (R + ). Then a and a sequence t n T such that u (t n ) H s (R + ) a. By the above existence argument applied at time t n for n sufficiently large, we obtain a contradiction, as follows. We shall select T = t n for n sufficiently large in a moment. We have, by assumption, u 1 Z solving the integral equation u 1 (t) = e it 2 x φ + L(f e i 2 x φ x=0 ) (19) λd(u 1 u 1 α 1 ) + λl(d(u 1 u 1 α 1 ) x=0 ) on [0, T ]. Apply the above existence argument to obtain u 2 Z solving, on 20

21 [T, T + δ], the integral equation u 2 (t) = e i(t T ) 2 x u(t ) + L T (f e i( T ) 2 x u(t ) x=0 ) (20) λd T (u 2 u 2 α 1 ) + λl T (D T (u 2 u 2 α 1 ) x=0 ) where L T g(t) = (g( + T ))(t T ) D T v(t) = D(v( + T ))(t T ) Since δ = δ(a, f 2s+1 ), we can select n sufficiently large so that T + δ = H (R + ) t n +δ > T. Now we show that we can concatenate these two integral equations. Define u(t) = u 1 (t) for < t T and u(t) = u 2 (t) for T t < +. Then clearly u L q t W r,s x C(R t ; H s x). Evaluate (19) at t = T, substitute into (20), and apply the two identities Lg(t) = e i(t T ) 2 x Lg(T ) L T (g e i( T ) 2 x Lg(T ) x=0 )(t) Dv(t) = e i(t T ) 2 x Dv(T ) + D T v(t) for t T for all t (21) with v(t) = λu u α 1 (t) and g(t) = f(t) e it 2 x φ x=0 Dv(0, t) on 0 t T +δ. This establishes that u solves, on [0, T + δ], the integral equation (18). Next, we show that u C(R x ; H 2s+1 ). Let ψ C such that ψ(t) = 0 for t 0, ψ(t) = 1 for T t T + δ, ψ(t) = 0 for t > T +δ. It is clear from the definition 2 2 of u that (1 ψ)u C(R x ; H 2s+1 t ). Since by (18) ψ(t)u(t) = ψ(t)e it 2 x φ + ψ(t)l(f θ2(t +δ) e i 2 x φ x=0 )(t) λψ(t)d(u u α 1 )(t) + λψ(t)l(θ 2(T +δ) D(u u α 1 ) x=0 ) by Lemmas 10(b), 11(b), and 1(b) we have ψu C(R x ; H 2s+1 t ). Next, we need to verify condition (2) in the definition of f. Now suppose u is a solution on [0, T u ] with T u T + δ, and v is a solution on [0, T v ] with 21

22 T v T + δ, and suppose min(t u, T v ) T. Then u(t) = v(t) for all t T (since T S). Then, again by (21), u solves (20) with u 2 replaced by u, and v solves (20) with u 2 replaced by v (u(t ) = v(t )). By uniqueness of the fixed point to (20) in L q [T,T +δ] W x s,r, we get that u(t) = v(t) on [T, T + δ]. We have thus established that sup S T + δ > T, which is a contradiction, so in fact lim t T u(, t) H s (R + ) = if T <. Now we move on the continuity claim. Suppose (φ, f) gives a solution u of (18) on [0, T ), and consider (φ 1, f 1 ) with φ φ 1 H s (R + ) + f f 1 H 2s+1 (R + ) < δ. Fix T < T. Let u 1 be the solution corresponding to (φ 1, f 1 ) on [0, T 1 ], where T 1 is the first time t such that u 1 L q [0,t] W x s,r = 2 u L q T W x s,r. We claim that T 1 > T provided we take δ sufficiently small. Indeed, taking the difference of the two integral equations, we find, for t min(t 1, T ) u u 1 L q [0,t] W x s,r cδ + c( u L q T W x s,r + u 1 L q T W s,r) u u q x 1 1 L 1 [0,t] Wx s,r where q 1 < q, and c depends only upon operator norms. This gives, by Lemma, u u 1 L q [0,t] W s,r x cδ (22) where now c depends on f, φ, and T. Now if T 1 < T, then take t = T 1 in (22) and δ sufficiently small to obtain a contradiction. The inequality (22) plus estimates on the difference of the integral equations for u and u 1 also shows u u 1 C([0,T ]; H s x ) + u u 1 C(Rx; H 2s+1 (0,T )) cδ Now we remark on the proof in the subcritical case (a) for 1 2 < s < 3 2. Let Z = C(R t ; Hx) s C(R x ; H 2s+1 t ) 22

23 Set r = 2, q = in the remainder of the argument above. Do note, however, that to show f(t) θ 2T (t)e it 2 x φ x=0 H 2s+1 (R + t ), we need to appeal to the compatibility condition f(0) = φ(0) and Lemma 3. Also, by Lemma 3, θ 2T (t)d(w w α 1 )(0, t) H 2s+1 0 (R + t ) q = Now we discuss the critical case (b). Let Z = L q t W s,r x with r = α+1 1+(α 1)s and (α+1). The integral equation is (α 1)(1 2s) Λw(t) = θ T (t)e it 2 x φ + θt (t)l(f θ 2T e i 2 x φ x=0 )(t) (23) λθ T (t)d(w w α 1 )(t) + λθ T (t)l(θ 2T D(w w α 1 ) x=0 )(t) Now, because q, θ T (t)e it 2 x φ L q t W x s,r 0 as T 0 and θ T (t)l(f θ 2T e i 2 x φ x=0 )(t) L q t W x r,s 0 as T 0. Therefore, T > 0 such that θ T (t)e it 2 x φ L q t W x s,r + θ T (t)l(f θ 2T e i 2 x φ x=0 )(t) L q t W x s,r < δ which gives Λw Z δ + c w α Z (2) For δ sufficiently small, there will be a fixed point in the space { w Z w Z < 2δ }. From Λu = u, (23) and Lemmas 10(a), 11(a), 1(a), we can recover the bounds in C(R t ; H s x), and by Lemmas 10(b), 11(b), 1(b), we can recover the bounds in C(R x ; H 2s+1 t ). Let T be the supremum of all existence times with a uniqueness stipulation, as before. We are not able to show the blowup statement in this case. Moreover, we also can only establish the continuity assertion for some T < T. 23

24 7.1 Notes on applying the chain and product rule We shall apply the chain rule (Lemma 5) with w : R C and F : C C given by F (w) = w α 1 w, for α 1. Then F (w) = (α 1) w α 3 (Re w) 2 + w α 1 (α 1) w α 3 (Re w)(im w) (α 1) w α 3 (Re w)(im w) (α 1) w α 3 (Im w) 2 + w α 1 and consequently each component of F (w) is bounded by w α 1. Thus where 1 r = 1 r 1 r = 1 2 r and 1 q so that 1 (α 1)r = 1 r s and 1 (α 1)q D s w α 1 w L r x cα w α 1 L r x Ds w L r x = 1 1 = 1 2. Since r, q have been selected q q q > 1, we have q D s w α 1 w L r x c Ds w α L r x To handle differences, for w 0, w 1 : R C, set w θ = θw 1 + (1 θ)w 0. Then w 1 α 1 w 1 w 0 α 1 w 0 = 1 θ=0 (α 1) w θ α 3 w θ (w θ (w 1 w 0 ))+ w θ α 1 (w 1 w 0 ) where z 1 z 2 = (Re z 1 )(Re z 2 ) + (Im z 1 )(Im z 2 ). To this, apply D s, and invoke the product rule (Lemma 6) and the chain rule (Lemma 5). 8 Uniqueness: Proof of Prop. 1 We shall begin by establishing uniqueness of a distributional solution with weak traces for the linear problem for s 0. Given two solutions u 1, u 2, consider the difference v = u 1 u 2. We are thus assuming v C([0, T ); L 2 (R + )) with v(x, 0) = 0 (25) 2

25 and lim v(x, ) x 0 + L 2 = 0 (26) (0,T ) Take T < T. Let θ(t) be a nonnegative smooth function supported on [ 2, 1] with θ = 1. Let θ δ (t) = δ 1 θ(δ 1 t). For δ, ɛ > 0, let v δ,ɛ (x, t) = v(y, s)θ δ (x y)θ ɛ (t s) dy ds (27) which defines, in the sense of distributions, v δ,ɛ (x, t) a smooth function on δ < x < +, ɛ < t < T 2ɛ. Owing to the assumption (25) we can write [ ] v δ,ɛ (x, t) = θ ɛ (t s) v(y, s)θ δ (x y) dy ds s where the integrals are defined in the usual sense. From this it follows that v δ,ɛ (, t) L 2 (R + x ) y sup v(, s) L 2 (R + x ) t+ɛ s t+2ɛ Owing to the assumption (26), L > 0 such that sup 0<x 2L v(x, ) L 2 (0,T ) 1. It follows that, for x + 2δ < 2L, (27) can be written [ ] v δ,ɛ (x, t) = θ δ (x y) v(y, s)θ ɛ (t s) ds dy s y where the integrals are understood in the usual sense, and we also have Let v δ,ɛ (x, ) L 2 (0,T ) sup v(y, ) L 2 (ɛ,t +2ɛ) (28) x+δ<y<x+2δ v ɛ (x, t) = θ ɛ (t s)v(x, s) ds s which is initially understood as defining, for each t, a distribution in x on (0, + ). It follows from (25) that it is also, for each t, a square integrable function in x with v ɛ (, t) L 2 (R + ) sup t+ɛ<s<t+2ɛ v(, s) L 2 (R + ) and lim v ɛ(, t) v(, t) L 2 ɛ 0 + (R + ) = 0 (29) 25

26 Now we proceed to the calculation. The identity is + 0 v δ,ɛ (x, T ) 2 dx = + x=0 v δ,ɛ (x, 0) 2 + 2Im T t=0 x v δ,ɛ (0, t)v δ,ɛ (0, t) dt (30) Now x 1 with 0 < x 1 < L such that x v δ,ɛ (x 1, t) = L 1 (v δ,ɛ (L, t) v δ,ɛ (0, t)), by the mean-value theorem. Again by the mean-value theorem, x 2 with 0 < x 2 < x 1 such that x v δ,ɛ (x 1, t) x v δ,ɛ (0, t) = x 1 2 xv δ,ɛ (x 2, t). Subtracting, x v δ,ɛ (0, ) L 2 (0,T ) L sup xv 2 δ,ɛ (y, ) L 2 (0,T ) + L 1 0 y L Bounding the terms on the right of this equation, we have sup 0 y L sup v δ,ɛ (x, ) L 2 (0,T ) sup v(x, ) L 2 (ɛ,t +2ɛ) 1 0<x<L δ<x<l+2δ v δ,ɛ (y, ) L 2 (0,T ) (31) We also have 2 xv δ,ɛ (x, t) = i t v δ,ɛ (x, t) = iɛ 1 θ δ (x y)(θ ) ɛ (t s)v(y, s) dy ds and thus sup 0<x<L 2 xv δ,ɛ (x, ) L 2 (0,T ) ɛ 1 sup δ<x<l+2δ v(x, ) L 2 (ɛ,t +2ɛ) ɛ 1 Hence, for fixed ɛ > 0 by Cauchy-Schwarz, bounding by (31) and (28), we have T Send δ 0 in (30) to get 0 + x=0 v δ,ɛ (0, t) x v δ,ɛ (0, t) dt 0 as δ 0 v ɛ (x, T ) 2 = and then send ɛ 0 and use (29). Now we prove Prop x=0 v ɛ (x, 0) 2 dx

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